Problem: Triangle ABC is an equilateral triangle and O is the center of its inscribed circle. If the area of the circle is $4\pi$ sq cm, what is the area, in square centimeters, of triangle ABC? Express your answer in simplest radical form.

Note: The area of a circle with radius $r$ is $\pi r^2.$
Explanation: First, we note that the radius of the inscribed circle is 2 cm (since $\pi r^2 = 4\pi$ implies that $r=2$, given that $r$ is nonnegative).

Let $X$ be the midpoint of side $BC$. Thus segment $OX$ is a radius of the inscribed circle: [asy]
unitsize(16);
draw(Circle((0,0),2));
draw(((-2*sqrt(3),-2)--(2*sqrt(3),-2)--(0,4)--cycle));
draw(((0,0)--(0,-2)));
draw(((-sqrt(3),1)--(0,0)--(sqrt(3),1)));
dot((0,4)); label("A",(0,4),N);
dot((-2*sqrt(3),-2)); label("B",(-2*sqrt(3),-2),SW);
dot((2*sqrt(3),-2)); label("C",(2*sqrt(3),-2),SE);
dot((0,0)); label("O",(0,0),N);
dot((0,-2)); label("X",(0,-2),S);
dot((-sqrt(3),1)); dot((sqrt(3),1));
label("2",(0,-1),E);
[/asy] Then $COX$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so the sides $OX$, $CX$, and $CO$ are in the ratio $1:\sqrt3:2$. Since $OX=2$, we have $CX=2\sqrt 3$.

Triangle $ACX$ is also a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so the sides $CX$, $AX$, and $AC$ are in the ratio $1:\sqrt3:2$. Thus, $AX=(2\sqrt3)(\sqrt 3)=6$.

Triangle $ABC$ has base $BC = 2(XC) = 4\sqrt 3$ and corresponding height $AX = 6$, so its area is $\frac{1}{2}(4\sqrt 3)(6) = \boxed{12\sqrt 3}$.